I have a Poisson process which generates $N$ events ($X_1,..X_N$) according to the Poisson distribution, with expectation value $\nu$. Each of those $X_n$ events has a $p$ probability of taking value $1$ and $1-p$ probability of taking value $0$. Consider $X$ to be the sum of all $X_n$ values:
$X=\sum_{n=1}^{N} X_n$
Intuitively speaking, I understand that the probability of generating X events with value 1 follows a Poisson distribution. Assuming $R$ to be the poisson rate ($\nu=RT$, where $T$ is observation time), it's obvious that the rate of events which return 1 is $Rp$, and therefore the mean of the Poisson distribution of events which generate 1 is $\nu p$. So the distribution looks like this:
$P(X) = \frac{e^{-\nu p}(\nu p)^X}{X!}$
But how do I prove it more rigorously? I started from the Poisson probability of generating N events $P(N | \nu) = \frac{e^{-\nu}\nu^N}{N!}$, and multiplied that with the binomial probability that $X$ of those events take value $1$, which is $P(X\ of\ those\ N\ are\ 1| N, p)=\frac{N!}{X!(N-X)!} p^X(1-p)^{N-X}$. What I end up with is:
$P(X)= \frac{(\nu p)^X}{X!} \cdot \frac{e^{-\nu}[\nu(1-p)]^{N-X}}{(N-X)!}$
How do I prove that the last factor in my RHS is $e^{-\nu p}$, such that it shows it's a Poisson process with expectation value $\nu p$? Are there some approximations I can use?